Let the function f(x)=(x^2-1)|x^2-ax+2|+ |x| be not differentiable at the two points x= =2…

Question

Let the function $f(x)=({x}^{2}-1)|{x}^{2}-ax+2|+\cos |x|$ be not differentiable at the two points $x=\alpha =2$ and $x=\beta $. Then the distance of the point $(\alpha ,\beta )$ from the line $12x+5y+10=0$ is equal to :

Accepted Answer

Since $\cos$ is an even function, $\cos|x|=\cos x$ is smooth everywhere; the only source of non-differentiability is $(x^2-1)|x^2-ax+2|$, at real roots of $x^2-ax+2=0$ (where $|x^2-ax+2|$ has a corner) that are not cancelled by a simultaneous zero of $(x^2-1)$.

Given one such point is $x=2$: substituting into $x^2-ax+2=0$ gives $4-2a+2=0\implies a=3$. Then $x^2-3x+2=(x-1)(x-2)$, with roots $x=1,2$.

At $x=2$: $x^2-1=3\ne0$, so this is a genuine non-differentiable point, consistent with $\alpha=2$. Comparing with the other root, $\beta=1$.

Distance from $(\alpha,\beta)=(2,1)$ to $12x+5y+10=0$:
$$d=\frac{|12(2)+5(1)+10|}{\sqrt{12^2+5^2}}=\frac{|24+5+10|}{13}=\frac{39}{13}=3$$

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